A copper wire at room temperature that is 1 mm thick and 30 cm long is connected to a 1 V battery. If all of the power goes into heating up the wire, find the temperature of the wire in 10s. (the density of copper is 8.9 g/cm3 and specific heat is 0.39 J/g·°C) I can't figure out this answer correctly. Thanks!!
ANZI said, ...T2=1925 'C Of course, if you try that experiment at home, the wire won't get anywhere near that hot. There are two reasons. One is the internal resistance of the battery. A real 1V battery would not be able to drive that many Amperes through the wire unless maybe it was around the size of an SUV. There would be less heat generated, and some of it would be generated within the battery itself. Reason number two is that the heat would not all build up in the wire. Much of it would be radiated away during the ten seconds that the current was applied.
the respond relies upon on a brilliant variety of factors. textile and length and thickness of the twine, and length and capability of the battery. If the twine gets too warm, it may react with the air. case in point, copper twine will turn green with copper oxide. Too warm and it will soften or set fireplace to this is environment. biking it on and stale should not be a difficulty, as long because it does not get too warm. Haw straight away it heats up and cools down (and that they are the comparable) relies upon, returned on the mass of the twine, the textile, and what this is touching. you will could test, start up wit a skinny twine, like #30, and a protracted length, possibly might up in a coil (you will could use magnet twine which has a layer of varnish on it) only as an occasion, in case you had a 9v battery and you wanted 3 watts of warmth, use P = E?/R 3 = 9?/R R = 27 ohms now use the resistivity formula, to calculate the dimensions Resistance of a twine in ? R = ?L/A ? is resistivity of the textile in ?-m L is length in meters A is circulate-sectional area in m? A = ?r?, r is radius of twine in m resistivity Cu 17.2e-9 ?-m #30 twine has a diameter of 0.25 mm or a radius of 0.000125 m it quite is a circulate-sectional area of A = ?r? = A = ?(0.000125)? = 4.9e-8 m? plugging into the resistivity formula 27 = (17.2e-9)L/(4.9e-8) L = 77 meters. this could be too long, so this is quite useful to objective for a thinner twine. Or swap to a twine with an more suitable resistivity. Nichrome, case in point, has a cost resistivity Nichrome 150e-8 ?-m 100x larger than copper. .
You also need the resistivity of copper which is p=1.68×10?8 ohm*m Lets make everything into meters d=1 mm =0,001 m l=30 cm =0,3 m U=1 V t=10s D=8900 kg/m''3 T=20 Deg C (room temperature) c=390 J/kg*C the area of the wire is A=pi*(d/2)''2=7,85*10''-7 m''2 ( n''2 n squared ) The Resistance of it is R=p*l/A R=6,4*10''-3 ohms Heat emitted is Q=t*U*I or Q=t*U''2/R Q=1558 J This heats the wire Q=c*m*(T2-T) where m=A*l*D So T2=Q/(c*A*l*D)+T T2=1925 'C
